angular mechanics - angular momentum contents: review linear and angular qtys angular vs. linear...

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Angular Mechanics - Angular Momentum Contents: Review Linear and angular Qtys Angular vs. Linear Formulas Angular Momentum Example | Whiteboard Conservation of Angular Momentum Example | Whiteboard

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Angular Mechanics - Angular Momentum

Contents:• Review

• Linear and angular Qtys• Angular vs. Linear Formulas

• Angular Momentum• Example | Whiteboard

• Conservation of Angular Momentum• Example | Whiteboard

Angular Mechanics - Angular Quantities

Linear:(m) s

(m/s) u(m/s) v

(m/s/s) a(s) t

(N) F (kg) m

(kgm/s) p

Angular: - Angle (Radians)

i - Initial angular velocity (Rad/s)

f - Final angular velocity (Rad/s)

- Angular acceleration (Rad/s/s)

t - Uh, time (s)

- Torque

I - Moment of inertia

TOCL - Angular momentum

Linear:s/t = vv/t = a

u + at = vut + 1/2at2 = su2 + 2as = v2

(u + v)t/2 = sma = F

1/2mv2 = Ekin

Fs = Wmv = p

Angular: = /t = /t* = o + t = ot + 1/2t2

2 = o2 + 2

= (o + )t/2* = IEk rot = 1/2I2

W = *

*Not in data packet TOC

L = I

Example: What is the angular momentum of a 23 cm radius 5.43 kg grinding wheel at 1500 RPMs?

TOC

Whiteboards: Angular momentum

1 | 2 | 3

TOC

What is the Angular Momentum of an object with an angular velocity of 12 rad/s, and a moment of inertia of 56 kgm2?

L = IL = (56 kgm2)(12 rad/s) = 672 kgm2/s

L = 670 kgm2/s

670 kgm2/s W

What must be the angular velocity of a flywheel that is a 22.4 kg, 54 cm radius cylinder to have 450 kgm2/s of angular momentum? hint

L = I, I = 1/2mr2

L = I = (1/2mr2) = L/(1/2mr2) = (450 kgm2/s)/(1/2(22.4 kg)(.54 m)2) = 137.79 rad/s = 140 rad/s

140 rad/s W

What is the angular momentum of a 3.45 kg, 33 cm radius bike wheel traveling 12.5 m/s. Assume it is a thin ring.

I = mr2 = (3.45 kg)(.33 m)2 = 0.3757 kgm2

= v/r = (12.5 m/s)/(.33 m) = 37.879 rad/sL = I = 14 kgm2/s

14 kgm2/s W

Example: A merry go round that is a 340. kg cylinder with a radius of 2.20 m has an 80.0 kg person who is 2.10 m from the center. If a torque of 94.0 mN acts for 15.0 s, what is the change in angular velocity of the merry go round?Ft = mΔvΓt = IΔω

Whiteboards: Torque, time, I and Δω

1 | 2 | 3

For what time does a torque of 12.0 mN need to be applied to a cylinder with a moment of inertia of 1.40 kgm2 so that its angular velocity increases by 145 rad/s

Γt = IΔω

16.9 s W

A grinding wheel that is a 5.60 kg 0.125 m radius cylinder goes from 152 rad/s to a halt in 22.0 seconds. What was the frictional torque?

Γt = IΔω

0.302 mN W

What is the mass of a cylindrical 2.30 m radius merry go round if we exert a force of 45.0 N tangentially at its edge for 32.0 seconds, it accelerates to a speed of 1.50 rad/s

Γt = IΔω

835 kg W

Angular Mechanics – Conservation of angular momentum

TOC

Conservation of Magnitude:• Figure skater pulls in arms

•I11 = I22• Demo

Example: A figure skater spinning at 3.20 rad/s pulls in their arms so that their moment of inertia goes from 5.80 kgm2 to 3.40 kgm2. What is their new rate of spin?What were their initial and final kinetic energies?(Where does the energy come from?)

Example: A merry go round is a 210 kg 2.56 m radius uniform cylinder. Three 60.0 kg children are initially at the edge, and the MGR is initially moving at 5.4 rad/s. What is the resulting angular velocity if they move to within 0.50 m of the center?

So Why Do You Speed Up?

A

B Concept 1B has a greater tangential velocity than A because of the tangential relationshipv = r

Whiteboards: Conservation of Angular Momentum

1 | 2 | 3

TOC

A gymnast with an angular velocity of 3.4 rad/s and a moment of inertia of 23.5 kgm2 tucks their body so that their new moment of inertia is 12.3 kgm2. What is their new angular velocity?

I11 = I22

(23.5 kgm2)(3.4 rad/s) = (12.3 kgm2)2

2 = (23.5 kgm2)(3.4 rad/s)/ (12.3 kgm2)2 = 6.495 = 6.5 rad/s

6.5 rad/s W

A 5.4 x 1030 kg star with a radius of 8.5 x 108 m and an angular velocity of 1.2 x 10-9 rad/s shrinks to a radius of 1350 m What is its new angular velocity? hint

I11 = I22 , I = 2/5mr2

2 = I11/I2 = (2/5mr12)1/(

2/5mr22)

2 = r121/r2

2

2 = (8.5 x 108 m)2(1.2 x 10-9 rad/s)/(1350 m)2

2 = 475.72 rad/s = 480 rad/s

480 rad/s W

A 12 kg point mass on a massless stick 42.0 cm long has a tangential velocity of 2.0 m/s. How fast is it going if it moves in to a distance of 2.0 cm? hint

I11 = I22 , = v/r, I = mr2

1 = v/r = (2.0 m/s)/(.420 m) = 4.7619 rad/sI1 = mr2 = (12 kg)(.42 m)2 = 2.1168 kgm2

I2 = mr2 = (12 kg)(.02 m)2 = .0048 kgm2

2=I11/I2=(2.1168 kgm2)(4.7619 rad/s)/(.0048 kgm2)

2= 2100 rad/s

2100 rad/s W

Angular Mechanics – Conservation of angular momentum

TOC

Angular momentum is a vector.(It has a Magnitude and a Direction)

Magnitude - I

Direction – Orientation of axis

TOC

Conservation of Angular momentum:

Magnitude• Planets around sun• Contracting Nebula | IP Demo• Crazy Merry go round tricks

Direction• Motorcycle

• On a jump• Revving (show drill) (BMW)

• People jumping from cliffs (video)• Aiming the Hubble• Demo stopping the gyroscope• Demo Turning a gyro over

Stability• Gyroscopes

• Demo small• Torpedoes• In a suitcase

• Demo hanging gyro• Bicycles